Download June 2007 - Vicphysics

2007 VCE Physics Unit 3 Exam Solutions (updated 18th June)
Consequentials are indicated as “Conseq on 1”
Suggested Marking Scheme in italics
These suggested solutions have been prepared by the AIP Education Committee. Every effort has
been made to double check the solutions for errors and typos.
Motion in one and two dimensions
1.
1.0 m
t = 0.45, u = 0, a = 10, s = ?, use s = ut+ ½ at2, s = 0 + ½ x 10 x (0.45)2 (1)
s = 1.01 m (1). Note: The use of 9.8 m/s2 gives 0.99 m to two significant figures.
2.
A parabolic path starting horizontal, but curving over (1) to meeting the ground at a point
immediately below the bike seat in position B (1). The horizontal velocity of the tail light keeps
its original value and travels forward with the bike as it falls. Note 1: This answer assumes that
Mary is stationary relative to the ground. Note 2: The vertical drop and the horizontal
displacement of the tail-light are no on the same scale.
3.
By conservation of momentum, momentum before = momentum after.
2 x 3 + 1 x 0 = 2 x v + 1 x 4 (1), solve for v (1).
4.
Elastic (1)
KE before = ½ x 2 x 32 = 9 J (1),
KE after = ½ x 2 x 12 + ½ x 1 x 42 = 1 + 9 = 9J (1)
5.
400 N
Average force = Change of momentum / Time taken (1)
Average force = (1 x 4 – 1 x 0) / 0.01 (1) = 400N (1).
6.
Zero
At 400 m air resistance equals the weight (1), so net force is zero (1).
2
7.
3.3 m/s
At 100m, AR = 400 N, so net force = 600 – 400 = 200 N (1).
Accel’n = 200/60 (1) = 3.3 m/s2 (1)
8.
250,000 J
Area under the graph from zero to 500m. 12.5 squares (1), each of 20,000 J gives
250,000 J (1). Note: The question is not well worded.
9.
0.40 kg
Extension = 0.6 – 0.4 = 0.2 (1) Mg = kx, find M. M x 10 = 20 x 0.20 (1),
M = 0.40 kg (1).
10. 0.50 J
Increase in PE = ½ kx22 - ½ kx12 (1) = ½ x 20 x ((0.30)2 – (0.20)2) (1)
Increase in PE = 0.50 J (1).
11. D
Total energy is conserved. (2)
12. A
Gravitational force = GMm/r2. G, M and m are same for both,
so ratio equals (RP/RE)2 = (6/10.5)2 = 0.33. (2)
13. 574 years
Using Kepler’s 3rd Law: R3/T2 is the same for both objects,
then (TE)2 = (RE/RP)3 x (TP)2 = (10.5/6)3 x 2482 (1) = 329623= 574 years. (1)
14. 3.4 s
u= 40sin250, v = -40sin250, a = -10, s = 0, t =?. Use v = u+ at
t = 2 x 40 x sin250 / 10 (1) = 3.38 s. (1)
Note 1: The use of 9.8 m/s2 gives the same answer to two significant figures.
Note 2: The data in the question is inconsistent. The time of flight can also be
determined from the range;
Time of flight = range / horizontal component of the speed
Time = 127 / 40cos250 which gives a different answer of 3.5 s.
15. 14.2 m
u= 40sin250, v = 0, a = -10, s = ?, t = 3.38 / 2 s (1), Use s = (u + v)t/2
s = 40sin250 x 3.38 / 4 (1) = 14.28 m (1) Conseq on Q’n 14.
16. D
Weight is only force and is always down. (2)
17. A, C
A: same max height so time to go up and down is unchanged (1). B: twice the
horizontal distance in the same time, so horizontal component is doubled, but
max height is the same so vertical component in unchanged. So the initial speed
is greater. C: acceleration equals g and unchanged (1). D: Horizontal component
is doubled, so angle is less.
Note: It could be argued that the question includes what happens in the gun where
the acceleration of the paintball is greater, so only giving A.
Electronics and photonics
1.
Peak voltage of 20 mV input becomes a 20 x 10-3 x 80 V peak output = 1.6 V (1). Graph is an
inverted sine wave (1) centred on zero with a peak voltage of 1.6 V.
2.
A peak voltage of 120 mV gives an output peak voltage of 9.6 V above and below the 3.0 volts
at Y. This means the output will be clipped at 6.0 V and at 0 V, which on the other side of the
capacitor will give a square looking wave (1) between +3 and -3 volts. (1)
Note 1: This answer assumes the value of the emitter resistor is zero and ignores the collector
emitter voltage.
Note 2: The frequency of the input signal has doubled as seen in Figure 3a, but it is not
mentioned in the text of the question.
3.
Saturation means maximum collector current (1). At this point the voltage drop across the
collector resistor is a maximum and so the voltage at the collector relative to the earth is a
minimum and cannot go any lower (1).This means the output voltage is clipped at the
bottom (1).
4.
B
All the answers have difficulties, but it is likely that B is the expected answer.
A: Wrong, an insulating medium is an insulator!, no movement of charge through
it, although it could be argued that there is a AC electric field passing through the
insulator. B: This answer confuses the operation of a capacitor as a decoupler or
smoother with that of a static filled capacitor. The capacitor is not connected
straight to an AC source, rather it is connected to the collector which is biased
and always positive. The capacitor allows small changes in the voltage at one
plate to be quickly passed to the other plate allowing AC to pass, but the actual
voltage drop across the capacitor is slows to change and so blocks DC. C:
Magnetic flux is Unit 4 content, this explanation would require a magnetisable,
conductive medium, which it is not. D: Wrong (2)
5.
500 ohms
At 10 mA, the voltage drop across the diode, from Fig 4, is 1.0 V (1), so there is
5.0 volts across the resistor (1). Using Ohm’s law,
its resistance is 5 / (10 x 10-3) = 500 ohms. (1)
6.
6.0 V (1)
Diode is reverse biased, so the left part of Fig 4 applies (1). The current is zero,
so the voltage drop across R is also zero and all the voltage is across
the diode (1).
7.
10 lux
From figure 6, when R = 3k ohms, so intensity = 10 lux. (2)
8.
750 ohms
From the figure 6, at 20 lux, Resistance of LDR = 1500 ohms (1). Using voltage
divider equation: 4 = 6 x (1500/(1500+R)), solve for R. R = 750 ohms. (1)
Alternatively voltage across LDR = 4V, so voltage across R = 2V, so R is half the
resistance of the LDR.
9.
decrease (1) Need: Vout = 4V for a higher lux value. Greater lux means smaller LDR
resistance, so R will also need to decrease to keep each resistor’s fraction of the
voltage the same. (1)
10. D
Only B and D refer to brightness. Brightness cannot go negative, so D. Note:
This answer assumes that the value at the origin of the graphs is zero and also
assumes that circuit for the laser diode is sufficiently biased so that the laser
diode still emits light even when the voltage at W goes negative. (2)
11. C
Only A and C refer to voltage. When the photodiode does not conduct the
voltage at Y would be constant, so A is not possible. Note: This answer assumes
that the value at the origin of the graphs is zero. (2)
Einstein’s relativity
1.
light/interference/no. It could be argued that since the experiment was investigating the velocity
of light relative to the earth, that either or both ‘Earth’ and ‘light’ could be considered. (1)(1)(1).
Note also that neither Michelson nor Morley (typo in the paper) thought the experiment
disproved the aether hypothesis.
2.
3.
4.
5.
6.
7.
8.
9.
10.
10 s
Speed of sound is reduced to 290 m/s (1). Distance equals 3.00km, so time
equals 3000/290 (1) = 10.3 s (1).
B
The distance as measured by the pilot is 90,000 km, but the speed of the signal is
unaffected by the speed of the spacecraft and is c, so the time will be 9 x 107 / (3
x 108) = 0.300 sec. (2)
1.005
Substitute v = 0.1000c into the formula for the Lorentz factor. (1) (1)
B
The length of the ship as measured by Ann is contracted or shortened,
L = L0/. (2)
No (1)
Both Ann and Bob observe a moving object (1). The direction is irrelevant.
Each will measure their moving object as shorter than its proper length. (1)
B
Moving clocks run slow. (2)
A
Moving clocks run slow. The life time of a moving pion will be longer that its
life time at rest. So a rest lifetime will be 4.16 x 10-7 / 16 = 2.60 x 10-8 s. (2)
See Q’n 4 substitute v = 0.998c into the formula for the Lorentz factor and show the answer is
close to 16. Actually it is 15.8. (1) (1) (1)
4.4 x 109 kg Use E = mc2 (1). m = 4.0 x 1026 / (3.0 x 108)2 (1) = 4.4 x 109 kg. (1)
Investigating materials and their use in structures
1.
length/shear/elasticity (1) (1) (1)
2.
Taking torques about T2: T1 x 3L/4 (1) = Mg x L/4 (1). Cancelling and rearranging (1)
T1 = Mg/3.
3.
2
Substitute Mg = 3T1 (1) into the supplied equation T2 = Mg – T1
gives T2 = 3T1 – T1 =2T1 (1)
4.
D
Between the cables the bottom is under tension, to the right of T2, the top is under
tension. (2)
5.
B
Both materials return to the same shape. (2)
6.
C
Cats iron is brittle, mild steel is ductile. (2)
4
7.
1.9 x 10 N Stress = Force/ Area (1). So Force = 240 x 106 x  x (10/2 x 10-3)2 (1)
Force = 1.9 x 104 N (1)
8.
3.8 J
Strain energy = Area under graph x Volume (1)
Strain energy = ½ x 240 x 106 x 0.4 x 10-3 x  x (10/2 x 10-3)2 x 1.0 (1) = 3.8 J (1)
9.
D
At B, weight of lamp is down, which is balanced by the upward component of the
force within the rod, which must therefore be under compression. The horizontal
and outward component of the compression force in the rod is balances by the
inward component of the force in the cable, which must therefore be under
tension. (2)
10. 120 N
upward component of the force in the rod = Weight. CB x sin600 = Mg (1)
Outward component of the force in the rod = Force in AB. CB x cos600 = AB (1)
Eliminating CB, AB = Mg / tan600 = 20 x 10 / tan 600 = 115.5 N (1)
Further electronics
1.
oscilloscope/undervoltage/capacitor. Although an audible hum could be argued for. (1) (1) (1)
2.
C
ratio of turns = 6.0/240, which is the same as 30/1200. (2)
3.
C
A: Wrong, current can go two ways at the top. B: Wrong, current cannot go
either way at the top. D: Wrong, current returns to the supply without going
through Vout. (2)
4.
B
full wave rectification. (2)
5.
400 F
Time constant ( = CR) is time to lose 63% of the voltage, which is about 40
milliseconds (1). R = 100 ohms, so C = 40 x 10-3 / 100 = 4.0 x 10-4
C = 400 x 10-6 F. (1)
6.
C
From Figure 1 in 20 ms the voltage across the capacitor will drop to 7.0 V, before
it rises again on the next half wave peak. C fits this description. (2)
7.
6.0 V
Zener diode is used in reverse bias mode, which is the graph in
8.
9.
10.
11.
12.
the third quadrant. (2)
Supply voltage of 9.0 V minus 6.0 V across Zener (1) equals 3.0 V across R. (1)
Conseq on Q’n 7
30 mA
Using V = IR, I = 3 / 100 = 30/1000 (1) = 30 mA. (1) Conseq on Q’n 8
With the whole class connected in parallel to the supply, the effective resistance is reduced (1).
This reduces the Time constant, increases the ripple (1) and decreases the effective DC voltage.
Increase the power supply voltage and put a 12V zener diode with a resistor in series across the
bridge rectifier (1). Increase the size of the capacitor. (1)
Trace should be a sine wave. The period, or time for one cycle, is 20 ms, which is 4 divisions
across the graph (1). The peak voltage is 4 x sqrt of 2 = 5.6 volts, which is just under three
divisions up the graph from the middle line, then the same amount below the middle line. (1)
3.0 V
Dot points not covered:
Motion is one and two dimensions
• analyse uniform circular motion of an object in a horizontal plane;
• analyse relative velocity of objects along a straight line and in two dimensions;
• distinguish between stationary (inertial) frames of reference and frames of reference that are
moving at constant speed relative to the stationary frame, including Galilean transformations in
one dimension between frames of reference;
• analyse transfers of energy between kinetic energy, potential energy and other forms of energy for
objects that
– undergo elastic and inelastic collisions
– move from position to position in a changing gravitational field, using only areas under forcedistance and field-distance graphs;
• analyse planetary and satellite motion modelled as uniform circular orbital motion in a universal
gravitation field, using
– a = v2/r = 42r/T2
Electronics and photonics
• describe energy transfers and transformations in electrical–optical, and optical–electrical
conversion systems using opto-electronic converters;
Einstein’s relativity
• describe Maxwell’s prediction that the speed of light depend sonly on the electrical and magnetic
properties of the medium it is passing through and not on the speed of the source or the speed of
the medium;
• contrast Maxwell’s prediction with the principles of Galilean relativity (no absolute frame of
reference; all velocity measurements are relative to the frame of reference);
• compare Einstein’s postulates and the assumptions of the Newtonian model;
• use simple thought experiments to show that
– the elapse of time occurs at different rates depending on the motion of an observer relative to an
event
– spatial measurements are different when measured in different frames of reference;
• explain the concepts of proper time and proper length as quantities that are measured in the frame
of reference in which objects are at rest;
• explain movement at speeds approaching the speed of light in terms of the postulates of Einstein’s
theory of relativity;
• explain the equivalence of work done into increased ‘mass energy’ according to Einstein’s
equation, E = mc2,
Investigating materials and their use in structures
• compare the tensile and compressive strengths and the stiffness or flexibility of different materials
under load to determine their suitability for use in structures such as columns, beams and arches;
• model the behaviour of materials under load in terms of extension and compression, graphically
and algebraically using Young’s modulus;
• use data to describe and predict the performance of a simple structure under load;
Further electronics
• select measuring devices for circuit analysis and faults diagnosis;
• select measurements of voltage and current (from the use of a multimeter and an oscilloscope) in
the DC power supply circuit to evaluate the operation of the circuit in terms of its design brief;
• calculate power dissipation in circuit elements;